Formulas for Higher-Order Derivatives
In general, to find n-th derivative of function $$$y={f{{\left({x}\right)}}}$$$ we need to find all derivatives of previous orders. But sometimes it is possible to obtain expression for n-th derivative that depends on $$$n$$$ and doesn't contain previous derivatives.
First of all it is easily to extend constant multiple, sum and difference rules.
Fact. For functions $$${f{{\left({x}\right)}}}$$$ and $$${g{{\left({x}\right)}}}$$$, and constant $$${c}$$$ we have that
- $$${{\left({c}{f}\right)}}^{{{\left({n}\right)}}}={c}\cdot{{f}}^{{{\left({n}\right)}}}$$$;
- $$${{\left({f{\pm}}{g}\right)}}^{{{\left({n}\right)}}}={{f}}^{{{\left({n}\right)}}}\pm{{g}}^{{{\left({n}\right)}}}$$$.
Now consider power function $$${y}={{x}}^{{a}}$$$.
We have that $$${y}'={a}{{x}}^{{{a}-{1}}}$$$, $$${y}''={a}{\left({a}-{1}\right)}{{x}}^{{{a}-{2}}}$$$.
In general, $$${\color{red}{{{{\left({{x}}^{{a}}\right)}}^{{{\left({n}\right)}}}={a}{\left({a}-{1}\right)}\ldots{\left({a}-{n}+{1}\right)}{{x}}^{{{a}-{n}}}}}}$$$.
Example 1. Find $$${{y}}^{{{\left({7}\right)}}}$$$ if $$${y}={2}{{x}}^{{5}}-{5}{{x}}^{{10}}$$$.
We first use constant multiple and sum rule and then formula for power function.
$$${{y}}^{{{\left({7}\right)}}}={{\left({2}{{x}}^{{5}}-{5}{{x}}^{{10}}\right)}}^{{{\left({7}\right)}}}={2}{{\left({{x}}^{{5}}\right)}}^{{{\left({7}\right)}}}-{5}{{\left({{x}}^{{10}}\right)}}^{{{\left({7}\right)}}}=$$$
$$$={2}{\left({5}\cdot{4}\cdot{3}\cdot{2}\cdot{1}\cdot{0}\cdot{\left(-{1}\right)}\cdot{{x}}^{{{5}-{7}}}\right)}-{5}{\left({10}\cdot{9}\cdot{8}\cdot{7}\cdot{6}\cdot{5}\cdot{4}\cdot{{x}}^{{{10}-{7}}}\right)}=-{3024000}{{x}}^{{3}}$$$.
For function $$${y}={{a}}^{{x}}$$$ we have that $$${y}'={{a}}^{{x}}{\ln{{\left({a}\right)}}}$$$, $$${y}''={{a}}^{{x}}{{\left({\ln{{\left({a}\right)}}}\right)}}^{{2}}$$$ and in general $$${\color{green}{{{{\left({{a}}^{{x}}\right)}}^{{{\left({n}\right)}}}={{a}}^{{x}}{{\left({\ln{{\left({a}\right)}}}\right)}}^{{n}}}}}$$$. In particular, $$${\color{blue}{{{{\left({{e}}^{{x}}\right)}}^{{{\left({n}\right)}}}={{e}}^{{x}}}}}$$$.
For function $$${y}={\sin{{\left({x}\right)}}}$$$ we have that $$${y}'={\cos{{\left({x}\right)}}}$$$, $$${y}''=-{\sin{{\left({x}\right)}}}$$$, $$${y}'''=-{\cos{{\left({x}\right)}}}$$$, $$${{y}}^{{{\left({4}\right)}}}={\sin{{\left({x}\right)}}}$$$.
Since $$${\cos{{\left({x}\right)}}}={\sin{{\left({x}+\frac{\pi}{{2}}\right)}}}$$$ then $$$\color{green} {{{{\left({\sin{{\left({x}\right)}}}\right)}}^{{{\left({n}\right)}}}={\sin{{\left({x}+{n}\frac{\pi}{{2}}\right)}}}}}$$$.
Similarly, $$$\color{purple} {{{{\left({\cos{{\left({x}\right)}}}\right)}}^{{{\left({n}\right)}}}={\cos{{\left({x}+{n}\frac{\pi}{{2}}\right)}}}}}$$$.
We saw that constant multiple and sum rules are easily extended for n-th derivative.
It is not a case with product rule.
So, let's see how product rule looks like in the case of higher-order derivatives.
Suppose that function $$${f{{\left({x}\right)}}}$$$ and $$${g{{\left({x}\right)}}}$$$ has finite derivatives up to n-th order.
We have that $$${\left({f{{g}}}\right)}'={f{'}}{g{+}}{f{{g{'}}}}$$$, $$${\left({f{{g}}}\right)}''={\left({f{'}}{g{+}}{f{{g{'}}}}\right)}'={\left({f{'}}{g}\right)}'+{\left({f{{g{'}}}}\right)}'={\left({f{''}}{g{+}}{f{'}}{g{'}}\right)}+{\left({f{'}}{g{'}}+{f{{g{''}}}}\right)}={f{''}}{g{+}}{2}{f{'}}{g{'}}+{f{{g{''}}}}$$$.
For third derivative we have $$${\left({f{{g}}}\right)}'''={\left({f{''}}{g{+}}{2}{f{'}}{g{'}}+{f{{g{''}}}}\right)}'={\left({f{''}}{g}\right)}'+{2}{\left({f{'}}{g{'}}\right)}'+{\left({f{{g{''}}}}\right)}'=$$$
$$$={\left({f{'''}}{g{+}}{f{''}}{g{'}}\right)}+{2}{\left({f{''}}{g{'}}+{f{'}}{g{''}}\right)}+{\left({f{'}}{g{''}}+{f{{g{'''}}}}\right)}={f{'''}}{g{+}}{3}{f{''}}{g{+}}{3}{f{'}}{g{''}}+{f{{g{'''}}}}$$$.
We see that these formulas are very similar to the expansion of binom of newton. Just instead of powers here are derivatives. If we denote $$${f{=}}{{f}}^{{{\left({0}\right)}}}$$$ and $$${g{=}}{{g}}^{{{\left({0}\right)}}}$$$ then we will obtain Leibniz Formula.
Leibniz Formula. $$${{\left({f{{g}}}\right)}}^{{{\left({n}\right)}}}={\sum_{{{i}={0}}}^{{n}}}{{C}_{{n}}^{{i}}}{{f}}^{{{\left({n}-{i}\right)}}}{{g}}^{{{\left({i}\right)}}}$$$.
We can extend this case when in product there are more than two functions, but this involves many formulas and calculations fairly complex.
Example 2. Find $$${{\left({{x}}^{{2}}{\cos{{\left({x}\right)}}}\right)}}^{{{\left({50}\right)}}}$$$.
Here $$${g{{\left({x}\right)}}}={{x}}^{{2}}$$$ and $$${f{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}$$$.
We have that, $$${{\left({f{{\left({x}\right)}}}\right)}}^{{{\left({k}\right)}}}={\cos{{\left({x}+{k}\frac{\pi}{{2}}\right)}}}$$$. Also $$${g{'}}{\left({x}\right)}={2}{x}$$$, $$${g{''}}{\left({x}\right)}={2}$$$, $$${g{'''}}{\left({x}\right)}={{g}}^{{{\left({4}\right)}}}{\left({x}\right)}=\ldots={0}$$$.
This means that in Leibniz formula all summands except first three will be equal 0. Thus, $$${{\left({f{{g}}}\right)}}^{{{\left({50}\right)}}}={{C}_{{50}}^{{0}}}{{f}}^{{{\left({50}\right)}}}{g{+}}{{C}_{{50}}^{{1}}}{{f}}^{{{\left({49}\right)}}}{g{'}}+{{C}_{{50}}^{{2}}}{{f}}^{{{\left({48}\right)}}}{g{''}}=$$$
$$$={1}\cdot{\cos{{\left({x}+{50}\frac{\pi}{{2}}\right)}}}\cdot{{x}}^{{2}}+{50}\cdot{\cos{{\left({x}+{49}\frac{\pi}{{2}}\right)}}}\cdot{2}{x}+\frac{{{50}\cdot{49}}}{{2}}\cdot{\cos{{\left({x}+{48}\frac{\pi}{{2}}\right)}}}\cdot{2}=$$$
$$$=-{{x}}^{{2}}{\cos{{\left({x}\right)}}}-{100}{x}{\sin{{\left({x}\right)}}}+{2450}{\cos{{\left({x}\right)}}}={\left({2450}-{{x}}^{{2}}\right)}{\cos{{\left({x}\right)}}}-{100}{x}{\sin{{\left({x}\right)}}}$$$.
So, $$${{\left({{x}}^{{2}}{\cos{{\left({x}\right)}}}\right)}}^{{{\left({50}\right)}}}={\left({2450}-{{x}}^{{2}}\right)}{\cos{{\left({x}\right)}}}-{100}{x}{\sin{{\left({x}\right)}}}$$$.